Determination of the largest gaps and tightness for a given fit.

1. Find the limit deviations for the hole and the shaft.

2. Graphically depict the landing.

3. Mark the main deviations on the image.

4. Determine and record the maximum dimensions of the hole and shaft.

5. Determine and write down the nature of the conjugation.

6. Write down the possible limit values for clearances and interferences.

7. Determine the clearance and (or) preload tolerance value.

Example III.

1. Find the limit deviations for the hole and the shaft.

For hole Æ 40 H7:

ES=+0.025mm

EI=+0 mm

For shaft Æ 40 f7:

es = – 0.025 mm

ei = – 0.050 mm

2. Graphically depict the landing.

3. Mark the main deviations on the image.

4. Find the limiting dimensions of the hole and shaft.

Largest hole diameter:

D max u003d D + ES u003d 40 + 0.025 u003d 40.025 (mm.)

Smallest hole diameter:

D min u003d D + EI u003d 40 + 0 u003d 40 (mm.)

Shaft diameter:

d max u003d d + es u003d 40-0.025 u003d 39.975 (mm.)

Smallest shaft diameter:

d min u003d d + ei u003d 40-0.050 u003d 39.950 (mm.)

5. The nature of the pairing – landing with a gap, because the smallest hole size limit is greater than the largest shaft size limit (the hole tolerance field is located above the shaft tolerance field)

6. Let’s write down the possible limiting values of the gaps.

smallest gap :

S min u003d D min – d max u003d EI – es u003d 40 – 39.975 u003d 0- (-0.025) u003d 0.025 (mm)

largest gap :

S max u003d D max – d min u003d ES – ei u003d 40.025 – 39.95 u003d 0.025-(-0.050) u003d 0.075 (mm)

6. Determine the clearance tolerance.

T s = S max – S min = (D max – d min ) – (D min – d max ) =(D max – D min )+( d max – d min )=T D +T d

T D u003d D max – D min u003d ES-EI u003d 40.025 – 40 u003d 0.025 (mm.)

T d u003d d max – d min u003d es-ei u003d 39.975 – 39.950 u003d -0.025 – (-0.050) u003d 0.025 (mm.)

T s u003d S max – S min u003d T D + T d u003d 0.075- 0.025 u003d 0.025 + 0.025 u003d 0.050 (mm)

Example IV.

1. Find the limit deviations for the hole and the shaft.

For bore Æ 100 H6: For shaft Æ 100 s7:

ES = + 0.022 mm es = + 0.106 mm

EI = + 0 mm ei = + 0.071 mm

2. Graphically depict the landing.

3. Mark the main deviations on the image.

+106

+71

basic deviation

N max N min shaft +0.071mm

+22

Æ100 basic hole deviation 0

4. Find the limiting dimensions of the hole and shaft.

Largest hole diameter:

D max u003d D + ES u003d 100 + 0.022 u003d 100.022 (mm.)

Smallest hole diameter:

D min u003d D + EI u003d 100 + 0 u003d 100 (mm.)

Shaft diameter:

d max u003d d + es u003d 100 + 0.106 u003d 100.106 (mm.)

Smallest shaft diameter:

d min u003d d + ei u003d 100 + 0.071 u003d 100.071 (mm.)

5. The nature of the pairing is an interference fit, because tightness in the connection is provided, i.e. the largest hole size limit is less than the smallest shaft size limit (the hole tolerance field is located under the shaft tolerance field).

6. Let’s write down the possible limiting values of interference:

maximum tightness

N max = d max – D min = es – EI = 100.106 – 100 = 0.106 – 0 = 0.106 (mm)

least tightness

N min = d min – D max = ei – ES = 10.071 – 100.022 = 0.071 – 0.022 = 0.049 (mm)

7. Preload tolerance

T N =N max -N min =(d max -D min )-(d min -D max )=(d max -d min )+(D max -D min )=T d + T D .

T D u003d D max – D min u003d ES-EI u003d 100.022 – 100 u003d 0.022 – 0 u003d 0.022 (mm.)

T d u003d d max – d min u003d es-ei u003d 100.106 – 100.071 u003d 0.106 – 0.071 u003d 0.035 (mm.)

T N u003d N max -N min u003d T d + T D .= 0.106-0.049 u003d 0.022 + 0.035 u003d 0.057 (mm.)

Example V.

1. Find the limit deviations for the hole and the shaft.

For hole Æ 15 H5: For shaft Æ 100 s7:

ES = + 0.008 mm es = + 0.004 mm

EI = + 0 mm ei = – 0.004 mm

2. Graphically depict the landing.

3. Mark the main deviations on the image.

4. Find the limiting dimensions of the hole and shaft.

Largest hole diameter:

D max u003d D + ES u003d 15 + 0.008 u003d 15.008 (mm.)

Smallest hole diameter:

D min u003d D + EI u003d 15 + 0 u003d 15 (mm.)

Shaft diameter:

d max u003d d + es u003d 15 + 0.004 u003d 15.004 (mm.)

Smallest shaft diameter:

d min u003d d + ei u003d 15-0.004 u003d 14.996 (mm.)

5. The nature of the mating is a transitional fit, in which it is possible to obtain both a gap and an interference, depending on the actual dimensions of the hole and the shaft (the tolerance fields of the hole and the shaft overlap).

6. Let’s write down the possible limiting values of clearances and interferences.

maximum tightness

N max u003d d max -D min u003d es – EI u003d 15.004 – 15 u003d 0.004 (mm)

largest gap

S max u003d D max -d min u003d ES – ei u003d 15.008 – 14.996 u003d 0.012 (mm)

7. Fit tolerance T N =T d +T D

T D u003d D max – D min u003d ES-EI u003d 15.008-15 u003d 0.008 – 0 u003d 0.008 (mm.)

T d u003d d max – d min u003d es-ei u003d 15.004 – 14.996 u003d 0.004 – (-0.004) u003d 0.008 (mm.)

T N u003d T d + T D u003d 0.008 + 0.008 u003d 0.016 (mm).

Task 3.

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